- A solution is a homogeneous mixture of one or more solutes dissolved in a solvent. - When a non-volatile solute is added to a solvent, it lowers the vapor pressure and raises the boiling point of the solvent. - Colligative properties, such as changes in vapor pressure, boiling point, and freezing point, depend only on the ratio of solute particles to solvent molecules and not on the chemical identity of the solute.

2.  In chemistry, a solution is
a homogeneous mixture composed of only
one phase. In such a mixture, a solute is a
substance dissolved in another substance,
known as a solvent. The solvent does the
dissolving. The soln. more or less takes on the
characteristics of the solvent including its phase,
and the solvent is commonly the major fraction
of the mixture.

3. CHARACTERISTICS:-
 A soln. is a homogeneous mixture.
 A soln. is a single phase system.
 The particles of solute in soln. can’t be seen by
naked eye.
 The soln. does not allow beam of light to scatter.
 A soln. is stable.
 The solute from the soln. can’t be separated
by filtration (or mechanically).

4. TYPES:-
1. Gas
 If the solvent is a gas, only gases are dissolved
under a given set of conditions. An example of a
gaseous soln. is air (oxygen and other gases dissolved
in nitrogen). Since interactions b/w molecules play
almost no role, dil. gases form rather trivial solns.
In part of the literature, they are not even classified
as solns., but addressed as mixtures.

5. 2. Liquid
If the solvent is a liquid, then gases, liquids, and solids
can be dissolved. Here are some examples:
 Gas in liquid:
 Oxygen in water
 Carbon dioxide in water – the soln. is accompanied
by a chemical rxn (formn. of ions). The visible
bubbles in carbonated water is not the dissolved gas,
but only an effervescence of carbon dioxide that has
come out of soln; the dissolved gas itself is not visible
since it is dissolved on a molecular level.

6.  Liquid in liquid:
 The mixing of two or more substances of the same
chemistry but diff conc.s to form a const.
(Homogenization of soln.s)
 Alcoholic beverages are basically soln.s of ethanol in
water.
 Solid in liquid:
 Sucrose (table sugar) in water
 Sodium chloride (table salt) or any other salt in
water, which forms an electrolyte: When dissolving,
salt dissociates into ions.

7. 3. Solid
 If the solvent is a solid, then gases, liquids and solids can be
dissolved.
 Gas in solids:
 Hydrogen dissolves rather well in metals, especially
in palladium.
 Liquid in solid:
 Mercury in gold, forming an amalgam
 Hexane in paraffin wax
 Solid in solid:
 Steel, basically a soln. of carbon atoms in a crystalline matrix
of iron atoms.
 Alloys like bronze and many others.
 Polymers containing plasticizers.

8. 1. Mass percentage
2. Volume percentage
3. Mass by volume percentage
4. Parts per million
5. Mole fraction
6. Molarity
7. Molality

9. 1. MASS PERCENTAGE
 Mass percentage is calculated as the mass of a
component div. by the total mass of the mixture,
multiplied by 100%.
 Also Known as: mass percent, (w/w)%
 Example:
Ordinary bleach is 5.25% NaOCl by mass,
which means each 100 g of bleach contains 5.25
g NaOCl.

10.  Mass Percent Formula is expressed as :-
 It is mathematically expressed as :-

11. 2. VOLUME-VOLUME PERCENTAGE
 Volume-Volume percentage (v/v %) is a
measure of conc. of a substance in soln. expressed as
the ratio of the volume of the soln. to the total volume
of the soln. multiplied by 100%.
 Examples:
Wine has a typical value of alcohol content v/v
% of 12%. This means there is 12 mL of ethanol
for every 100 mL of wine.

12.  Volume Percent Formula is expressed as :-

13. 3. WEIGHT/VOLUME PERCENT
 Another variation on percentage conc. is
weight/volume percent or mass/volume percent.
This variation measures the amount of solute in
grams but measures the amount of soln. in millilitres.
 An example would be a 5%(w/v) NaCl
solution. It contains 5 g of NaCl for every
100. mL of solution.

15. 4. PARTS PER MILLION
 When the solute is present in trace quantities, it
is convenient to express the conc. in parts per
million (ppm).
 It is defined as the quantity of the solute in
grams present in 106 grams of the soln.
 Example:-
10 ppm of SO2 in air means 10 mL of
SO2 is present in 106 mL of air.

16. 

17. 5. MOLE FRACTION
 This method is used when the soln. is constituted by
mixing two or more components.
 It is defined as the ratio of no. of moles of one
component to the total no of moles of the soln. (i.e., all
the components).
 In a binary soln.,
Mole fraction of solute + Mole fraction of solvent =
1

19. 6. MOLARITY(Molar Concentration)
 It is defined as the no. of moles of the solute per
litre or per dm3 of the soln.
 Let wA g of the solute of molecular mass
mA be dissolved in V litre of soln.
Molarity of the soln. = wA/mA×V
 The unit of molarity is mol litre−1 or mol dm−3

21. 7. MOLALITY
 It is defined as the no. of the moles of the
solute present in 1 kg of the solvent, It is
denoted by m.
 Let wA grams of the solute of molecular
mass mA be present in wB grams of the
solvent, then,

24.  According to this law at a const. temp., the
amount of a given gas that dissolves in a given
type and volume of liquid is directly
proportional to the partial pressure of that gas
in equilibrium with that liquid.
» The solubility of a gas in a liquid is directly
proportional to the partial pressure of the gas
above the liquid.

25.  Henry's law can be put into mathematical terms (at
constant temperature) as:-
where p is the partial pressure of the solute in the gas
above the soln., c is the conc. of the solute and kH is
a constant with the dimensions of pressure divided by
conc.
The const., known as the Henry's law const.,
depends on the solute, the solvent and the temp.

27.  Solubility is the property of a solid, liquid,
or gaseous chemical
substance called solute to dissolve in a solid, liquid,
or gaseous solvent to form a homogeneous soln. of
the solute in the solvent.
 The solubility of a substance fundamentally
depends on the physical and chemical properties of
the used solute and solvent as well as on temp.,
pressure and the pH of the soln.

28. FACTORS AFFECTING
SOLUBILITY
1. Temperature:- The solubility of a given solute in
a given solvent typically depends on temp. As the
temp. of a liquid increases, the solubilities of gases
in that liquid decrease and vice versa.
2. Pressure :- For condensed phases (solids and
liquids), the pressure dependence of solubility is
typically weak and usually neglected in practice.

29. SOLUBILITY OF GASES
 Henry's law is used to quantify the solubility
of gases in solvents. The solubility of a gas in
a solvent is directly proportional to the partial
pressure of that gas above the solvent. This
relationship is written as:

31.  According to this law, the partial pressure of any
volatile constituent of a soln. at a const. temp. is
equal to the vapour pressure of pure constituent
multiplied by the mole fraction of that constituent in
the soln.
 Let a mixture (soln.) be prepared by mixing
nA moles of liquid A and nB moles of liquid B.
Let pA and pB be the partial pressures of two
constituents A and B in soln. and pA
0 and
pB
0 the vapour pressures in pure state respectively.

32.  Thus, according to Raoult’s law,
pA = nA/nA+nB pA
0 = mole
fraction of A pA
0 = XApA
0
And pB = nB/nA+nB pA
0 = mole
fraction of B pB
0 = XBpB
0
If the total pressure be P, then
P = pA + pB
= nA/nA+nB pA
0 + nB/nA+nB pA
0
= XAPA
0 + XBPB
0

35. Ideal solutions
Non-ideal solutions
Positive deviation from
Raoult’s law
Negative deviation
from Raoult’s law
1. Obey Raoult’s law at
every range of
concentration.
1. Do not obey Raoult’s
law.
1. Do not obey Raoult’s
law.
2. Δ Hmix = 0;
is neither evolved nor
absorbed during
dissolution.
2. Δ Hmix>0.
Endothermic dissolution;
heat is absorbed.
2. Δ Hmix<0.
Exothermic
dissolution; heat is
evolved.
3. ΔVmix = 0;
total volume of
solution is equal to
sum of volumes of the
components.
3. ΔVmix > 0.
Volume is increased after
dissolution.
3. ΔVmix <0.
Volume is decreased
during dissolution.
4. A—A,A—B, B—B
interactions should be
same
4. A—B attractive force
should be weaker than
A—A and B—B attractive
4. A—B attractive force
should be greater
than A—A and B—B

36. Examples:
Dil. solns;
C6H6 + toluene;
n-C6H14 + n-C7H16;
chlorobenzene +
bromobenzene;
n-butyl chloride + n-
butyl bromide.
Examples:
oacetone +
C2H5OH;
oacetone + CS2;
oH2O + CH3OH;
oH2O + C2H5OH;
oCCl4 + toluene;
oCCl4 + CHCl3;
oacetone + C6H6;
oCCl4 + CH3OH;
oCyclohexane +
C2H5OH
Examples:
oacetone + aniline;
oacetone + CCl3;
oCH3OH + CH3COOH;
oH2O + HNO3;
oCCl3 + diethyl ether,
oH2O + HCl;
oCH3COOH+ pyridine;
oCCl3 + C6H6

39. 1. VAPOUR PRESSURE OF
LIQUID-LIQUID SOLUTIONS
 If P = total vapour pressure
 pA = partial vapour pressure of component 1.
 pB = partial vapour pressure of component 2.
 The quantitative relationship b/w the partial
pressures ( pA , pB ) and the corresponding
mole fractions (XA, XB ) of the two components 1
& 2 respectively, was given by “Marte
Raoult” known as Raoult’s law

40. FROM DALTON’S LAW OF
PARTIAL PRESSURE

41. The plot shows that the graph of pA &,pB verses the mole fractions XA
&XB fora soln. is linearhence pA &,pB are directlyproportionalto XA
&XB respectively.
In Figure, the dashed lines I & II represent the partial
pressures of components, while the marked Line III
represents the total vapour pressure of the soln.

42. 2. VAPOUR PRESSURE OF
SOLUTIONS OF SOLIDS IN LIQUIDS
 In a pure liquid the entire surface is filled by only
liquid molecules, but when we add a non-volatile solute
to a liquid soln., a part of the liquid surface is also
occupied by these solutes. This reduces the space to
the liquid molecules to evaporate out resulting the
lowering in vapour pressure because only solvent part
of the soln. can vaporise and creates vapour pressure,
but not the solute part being non-volatile.

43. 1. Relative lowering of vapour
pressure
2. Elevation of boiling point
3. Depression of freezing point
4. Osmotic pressure.

44.  The word Colligative is derived from the
Latin ’Colligatus’ meaning bound together.
 This no. can be related to the various units
for conc. of solns.
Colligative properties are mostly studied for dil.
soln.s, whose behaviour may often be
approximated as that of an ideal soln.

45. In chemistry, Colligative Properties are
properties of soln.s that depend upon the ratio of the
no. of solute particles to the no. of solvent molecules
in a soln.
We will only consider those properties which result
because of dissolution of non -volatile solute in a
volatile liquid solvent.
They are independent of the nature of the solute
particles, and are due to the dil.n of the solvent by
the solute.

46. 1. RELATIVE LOWERING
OF VAPOUR PRESSURE
When a non-volatile solute is added to a solvent,
the vapour pressure is lowered due to the foll. reason:
Percentage surface area occupied by the solvent
decreases. Thus the rate of evaporation and vapour
pressure decreases. The solute molecules occupy the
surface, and so the per cent surface area occupied by
the solvent decreases.

47. Derivation Of Raoult’s Law
For A Dilute Solution
 It states that the relative lowering in vapour
pressure of a dil. soln. is equal to mole fraction
of the solute present in the soln.
 If n moles of solute be dissolved in N moles
of the solvent, the mole fraction of the solute will
be n/n+N.

48.  According to Raoult’s law, [p0–ps/p0] =
n/n+N.
This is the mathematical expression for modified form
of Raoult’s law. The above r/l can be written as,
p0/P0–ps = n+N/n = 1 + N/n
or p0/p0–ps – 1 = N/n or p0/p0–ps= N/n
or p0–ps/ps = n/N = wA/mA mB/wB

50. 2. ELEVATION OF BOILING
POINT
The b. pt. of a liquid is the temp. at which its
vapour pressure is equal to the atmospheric pressure.
The vapour pressure of a liquid is lowered when a
non-volatile solute is added to it. Hence, the b. pt.
of the solvent is elevated by the addition of non-
volatile solute.
 The difference in the b. pt. of the soln. & the b. pt.
of the pure solvent is termed elevation of b. pt.

51.  Elevation of b. pt., (ΔT) = B. pt. of the
soln. – B. pt. of pure solvent
 P0C = atmospheric pressure
 T0, T1 and T2= b. pts. of pure solvent,
soln. I and soln. II respectively
 P0, P1 and P2 = vapour pressure of pure
solvent, soln. I and soln. II at temp.
T0 respectively

52. T2–T0/T1–T0 = P0–P2/P0–P1
or ΔT2/ΔT1 = ΔP2/ΔP1
or ΔT ∝ ΔP
From Raoult’s law for dil. soln.
p0–ps/p0 = wA/mA.mB/wB
(ps = vapour pressure of soln.)
or p0–ps = wA/mA.mB/wB . p0
For the pure solvent, P0 (its vapour
pressure at the b. pt.) and mB (its
molecular mass) are const.

53. Therefore,
p0 – ps ∝ wA/mAwB
or Δp ∝ ΔT ∝ wA/mAwB
orΔT =K.wA/mAwB …(i)
where K is a const., c/a as elevation
const.
If wA/mA and wB = 1000 g.
ΔT = K/1000 = Kb

54. Kb is c/a molal elevation const. It is
defined as the elevation in b. pt.
produced when 1 mol of the solute is
dissolved in 1000 g of the solvent.
Thus, K = 1000 Kb
Putting this value in Eq. (i),
ΔT° = 1000 Kb wA/mAwB …(ii)
or ΔT = Molality × Kb
(since wA/mA×wB × 1000 = molality).

56. 3. DEPRESSION OF FREEZING
POINT
Freezing point of a substance is defined as the temp. at
which the vapour pressure of its liquid is equal to the
vapour pressure of the corresponding solid.
Since the addition of a non-volatile solute always lowers
the vapour pressure of solvent, therefore, it will be in
equilibrium with solid phase at a lower pressure and hence
at a lower temp.
The difference b/w the f. pts. of the pure solvent and its
soln, is c/a depression of f. pt.

57.  Depression of f. pt. (ΔT) = F. pt. of the
solvent – F. pt. of the soln.
 CFB is a curve for solid solvent.
 The solvent, soln.I and soln.II vapour
pressure curves meet CFB curve at pts.B, F
and C respectively.
 Thus, T0, T1 and T2 for solvent, soln. I and
soln. II are, thus, P0, P1 and P2 respectively.

58. T0–T2/T0–T1 = P0–P2/P0–P1
or ΔT2/ΔT1 = ΔP1/ΔP2 or ΔT ∝ ΔP
From Raoult’s law for dil. soln.s,
p0–ps/p0 = wA/mA . mB/wB
or p0–ps = wA/mA . mB/wB . P0
For the pure solvent, p0 and mB are const.
Therefore,
p0 – ps ∝ wA/mAwB
or Δp ∝ wA/mAwB
or Δp ∝ ΔT ∝ wA/mAwB
or ΔT =K. wA/mAwB …(i)
where K is a const., c/a depression const.

59. When wA/mA (or mole of solute) and
wB = 1 g
ΔT = K
If wA/mA and wB = 100 g,
ΔT = K/100 = K'
Thus, K = 100 K’
Putting this value in Eq. (i),
ΔT = 100K'×wA/mA×wB …(ii)
If wA/mA and wB = 1000 g
ΔT = K/1000 = Kf

60. Kf is c/a molal depression const. It is
defined as the depression of f. pt. produced
when 1 mole of solute is dissolved in 1000 g
of the solvent.
Thus, K = 1000Kf
Putting this value in Eq. (i),
ΔT° = 1000
Kf wA/mAwB …(iii)
or ΔT = molality × Kf

63. 4. OSMOTIC PRESSURE.
OSMOSIS: When a semi-permeable
membrane is placed b/w a soln. and a solvent, it is
observed that solvent molecules enter the soln. and
the volume of soln. increases.
It is also observed that if two soln.s of unequal
conc.s are separated by a semi permeable membrane,
the solvent molecules from a soln. of lower conc. move
towards a soln. of higher conc.

66. Osmotic Pressure is the pressure which
needs to be applied to a soln. to prevent the
inward flow of water across a semi permeable
membrane.
The phenomenon of osmotic pressure arises
from the tendency of a pure solvent to move
through a semi-permeable membrane and into a
soln. containing a solute to which the
membrane is impermeable.

68. By Van't Hoff's law, Osmotic Pressure
Formula can be calculated as:
Where ,
i = number of ions formed by dissociation of a
solute molecule
R = ideal gas const.
T = absolute temp. (measured in K)
c = molal or molar conc.

69. The expression which relates osmotic pressure to conc. &
temp. is similar to the ideal gas eqn:
The Osmotic Pressure Formula of an ideal soln. can be
calculated as follows:
Where ,
P = osmotic pressure in mm Hg
n = no. of solute particles
c/M = molar conc. of the substance in mol/kg
R = universal gas const.
T = absolute temp.

70. TYPES OF SOLUTIONS BASED
ON SOLUTE CONCENTRATION
 The two soln.s are equal in their solute concs.
We say that they are isotonic to each other.

71. The soln. in the bag contains less solute than the
soln. in the beaker. The soln. in the bag is hypotonic
(lower solute conc.) to the soln. in the beaker. The
soln. in the beaker is hypertonic (higher solute conc.)
to the one in the bag. Water will move from the
hypotonic soln. into the hypertonic soln.

72. REVERSE OSMOSIS
Reverse osmosis (RO) is a water purification
technology that uses a semi permeable membrane.
In RO, an applied pressure is used to
overcome osmotic pressure, that is driven by chemical
potential. RO can remove many types
of molecules and ions from solutions and is used in
both industrial processes and to produce potable
water.

73. The result is that the solute is retained on the
pressurized side of the membrane and the
pure solvent is allowed to pass to the other side.
To be "selective," this membrane should not allow
large molecules or ions through the pores (holes),
but should allow smaller components of the solution
(such as the solvent) to pass freely.
Reverse osmosis is most commonly known for its
use in drinking water purification from seawater,
removing the salt and other effluent materials from
the water molecules.